Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define FASTIO

#ifdef FASTIO
#define scanf abacaba
#define printf abacaba
#endif

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/
const int MAXN = 210000;

int n;
int p[MAXN];
int pos[MAXN];

int main() {
#ifdef FASTIO
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
	cin >> n;
	for (int i = 0; i < n; ++i) {
		cin >> p[i];
		--p[i];
		pos[p[i]] = i;
	}
	int ans = 0;
	int lst = n - 1;
	for (int i = n - 2; i >= 0; --i) {
		if (ans == 0) {
			if (pos[i] > pos[i + 1]) {
				lst = i + 1;
				++ans;
			}
		}
		else {
			int a = pos[lst];
			int b = pos[i];
			int c = pos[i + 1];
			if ((a < b && b < c) || (c < a && a < b) || (b < c && c < a)) {
				continue;
			}
			else {
				++ans;
				lst = i + 1;
			}
		}
	}
	cout << ans << "\n";
	return 0;
}